---

Let $B\in M_n(\mathbb Z)$ be the nonsingular upper-triangular matrix from [[additive_combinatorics/adamczewski_2026_erdos1/lattice_reduction|the lattice reduction]]. Let $S$ be the $(n+1)\times(n+1)$ matrix whose first column and last row vanish and whose block in the first $n$ rows and last $n$ columns is $B$. Because $B$ is upper triangular, every nonzero entry of $S$ lies strictly above the diagonal. For an integer $t$, put

$$U_t=I+tS.$$

This is an integer upper-unitriangular matrix with determinant $1$.

Define the unimodular balancing map

$$\beta(x_0,\ldots,x_n) =(x_0,\ldots,x_{n-1},x_n-x_0-\cdots-x_{n-1})$$

and, for $z\in\mathbb Z^n$,

$$\Phi_t(z)=\beta U_t(0,z).$$

Direct multiplication gives

$$\Phi_t(z)=tL_B(z)+E(z),\qquad L_B(z)=\left(Bz,-\sum_i(Bz)_i\right),\quad E(z)=\beta(0,z). \tag{1}$$

Put

$$A(B)=\sum_{i,j}|\operatorname{adj}(B)_{ij}|,\qquad E_B=(n+1)A(B).$$

Statement

The error term is dominated by the balanced norm: each $z\in\mathbb Z^n$ satisfies

$$\|E(z)\|_\infty\leq E_B\|L_B(z)\|_\infty.$$

Proof

Put $M=\|L_B(z)\|_\infty$, so that $\|Bz\|_\infty\leq M$ by the definition of $L_B$. The integer $\det B$ is nonzero, so $|\det B|\geq1$, and

$$(\det B)z=\operatorname{adj}(B)Bz.$$

Consequently every $|z_i|\leq A(B)M$. The first $n$ coordinates of $E(z)=\beta(0,z)$ are drawn from $0,z_1,\ldots,z_{n-1}$, while its last coordinate is $z_n-z_1-\cdots-z_{n-1}$. Each is bounded in absolute value by $(n+1)A(B)M=E_BM$.

Source and dependencies

An explanation of the proof of Erdős Problem 1, preliminary exposition with no named author (erdosproblems.com, 2026), §5, equations (17)–(19) and Lemma 5.1, pp. 6–7. The edition read is named on the source card. The matrix identities are exact; the estimate uses only the adjugate identity.

Bears on. #1.